To find all solutions to the equation [tex]\(\cot \theta + \sqrt{3} = 0\)[/tex] within the interval [tex]\([0, 2\pi)\)[/tex], we can follow these steps:
1. Rewrite the equation:
[tex]\[
\cot \theta + \sqrt{3} = 0
\][/tex]
This implies:
[tex]\[
\cot \theta = -\sqrt{3}
\][/tex]
2. Express cotangent in terms of tangent:
Recall that [tex]\(\cot \theta = \frac{1}{\tan \theta}\)[/tex]. So, we have:
[tex]\[
\frac{1}{\tan \theta} = -\sqrt{3}
\][/tex]
Taking the reciprocal of both sides, we get:
[tex]\[
\tan \theta = -\frac{1}{\sqrt{3}}
\][/tex]
We can simplify [tex]\(\tan \theta\)[/tex] further as:
[tex]\[
\tan \theta = -\frac{\sqrt{3}}{3}
\][/tex]
3. Determine the reference angle:
We know that [tex]\(\tan \theta = \frac{\sqrt{3}}{3}\)[/tex] for [tex]\(\theta = \frac{\pi}{6}\)[/tex]. Therefore, for [tex]\(\tan \theta = -\frac{\sqrt{3}}{3}\)[/tex], the reference angle is still [tex]\(\frac{\pi}{6}\)[/tex], but the tangent function takes negative values in the second and fourth quadrants.
4. Find the solutions in the specified interval:
- Second Quadrant:
[tex]\[
\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
\][/tex]
- Fourth Quadrant:
[tex]\[
\theta = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{1\pi}{6} = \frac{11\pi}{6}
\][/tex]
Thus, the solutions to the equation [tex]\(\cot \theta + \sqrt{3} = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[
\theta = \frac{5\pi}{6}, \frac{11\pi}{6}
\][/tex]
In decimal form, these are approximately [tex]\(2.6179938779914944\)[/tex] and [tex]\(5.759586531581287\)[/tex], respectively, which confirms our earlier findings.
So, the final answer is:
[tex]\[
\theta = \frac{5\pi}{6}, \frac{11\pi}{6}
\][/tex]
